import unittest
from datetime import datetime

class GetFreqNumbersFromList(unittest.TestCase):
 def setUp(self):
  print("\n")
  self.start_time = datetime.now()
  print(f"{self._testMethodName} start: {self.start_time}")

 def tearDown(self):
  self.end_time = datetime.now()
  print(f"{self._testMethodName} end: {self.end_time}")
  exec_time = (self.end_time - self.start_time).microseconds
  print(f"{self._testMethodName} exec_time: {exec_time}")

 def normal_solution(self, _list, _debug=False):
  """
  普遍性解法
  利用字典記錄每個(gè)元素出現(xiàn)的次數(shù)——然后找出元素出現(xiàn)次數(shù)超過(guò)數(shù)組長(zhǎng)度一半的元素
  普遍性解法針對(duì)任何次數(shù)的統(tǒng)計(jì)均適用而不光只是針對(duì)出現(xiàn)次數(shù)超過(guò)數(shù)組長(zhǎng)度一半的情況
  """
  _target = len(_list) // 2
  _dict = {}
  for _member in _list:
   if _member not in _dict:
    _dict.setdefault(_member, 1)
   else:
    _dict[_member] += 1
  _ret = [_member for _member in _dict if _dict[_member] > _target]
  if _debug:
   print(_ret)
  return _ret

 def specific_solution(self, _list, _debug=False):
  """
  特殊性解法
  假設(shè)有兩個(gè)元素出現(xiàn)的次數(shù)都超過(guò)數(shù)組長(zhǎng)度一半就會(huì)得出兩個(gè)元素出現(xiàn)的次數(shù)超出了數(shù)組長(zhǎng)度的矛盾結(jié)果——所以超過(guò)數(shù)組長(zhǎng)度一半的元素是唯一的
  排序后在數(shù)組中間的一定是目標(biāo)解
  特殊性解法只能針對(duì)元素出現(xiàn)次數(shù)超過(guò)數(shù)組長(zhǎng)度一半的情況
  """
  _list.sort()
  if _debug:
   print(_list[len(_list) // 2])
  return _list[len(_list) // 2]

 def test_normal_solution(self):
  actual_result = self.normal_solution([2,2,2,2,2,2,1,1,1,1,1], False)
  self.assertEqual(actual_result[0], 2)

 def test_specific_solution(self):
  actual_result = self.specific_solution([2,2,2,2,2,2,1,1,1,1,1], False)
  self.assertEqual(actual_result, 2)

if __name__ == "__main__":
 # 找出出現(xiàn)次數(shù)超過(guò)數(shù)組長(zhǎng)度一半的元素
 suite = unittest.TestSuite()
 suite.addTest(GetFreqNumbersFromList('test_normal_solution'))
 suite.addTest(GetFreqNumbersFromList('test_specific_solution'))
 runner = unittest.TextTestRunner()
 runner.run(suite)

from math import pow, sqrt

def calc_circle_s_with(r, dy, x_slices):
  x_from_start_to_cc = sqrt(1 - pow(dy, 2))
  dx = x_from_start_to_cc / x_slices
  x_to_edge = 1 - x_from_start_to_cc
  quarter_circle_s = 0
  while x_to_edge < 1:
    rect_s = dy * dx
    quarter_circle_s += rect_s
    x_to_edge = x_to_edge + dx
    dy = sqrt(1 - pow((1 - x_to_edge), 2))
  circle_s = 4 * quarter_circle_s
  print(circle_s)

calc_circle_s_with(1, 0.0001, 10000000)